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Chapter 2: Problem 1

In Problems \(1-10,\) find the functions \(f+g, f-g, f g\), and \(f / g\), and give their domains. $$ f(x)=x^{2}+1, g(x)=2 x^{2}-x $$

Short Answer

Expert verified
Add: \((f+g)(x) = 3x^2 - x + 1\); Subtract: \((f-g)(x) = -x^2 + x + 1\); Multiply: \((fg)(x) = 2x^4 - x^3 + 2x^2 - x\); Divide: \((f/g)(x) = \frac{x^2 + 1}{2x^2 - x}\); Domain \(f/g: \mathbb{R} \setminus \{0, 0.5\}\).

Step by step solution

01

Add the Functions

To find \( (f+g)(x) \), add the two functions. Start with \( f(x) = x^2 + 1 \) and \( g(x) = 2x^2 - x \). Thus,\[(f+g)(x) = (x^2 + 1) + (2x^2 - x) = 3x^2 - x + 1\].
02

Subtract the Functions

To find \( (f-g)(x) \), subtract \( g(x) \) from \( f(x) \). Thus,\[(f-g)(x) = (x^2 + 1) - (2x^2 - x) = x^2 + 1 - 2x^2 + x = -x^2 + x + 1\].
03

Multiply the Functions

To find \( (fg)(x) \), multiply \( f(x) \) and \( g(x) \) together. Calculate as follows: \( (x^2 + 1)(2x^2 - x) \). Expanding this, \[ fg(x) = 2x^4 - x^3 + 2x^2 - x\].
04

Divide the Functions

To find \( \left( \frac{f}{g} \right)(x) \), divide \( f(x) \) by \( g(x) \). So, \[ \left(\frac{f}{g}\right)(x) = \frac{x^2 + 1}{2x^2 - x} \].
05

Determine the Domains

The domain of \( f(x) \) and \( g(x) \) both is all real numbers, \( \mathbb{R} \). Thus, the domains of \( f+g, f-g, \) and \( fg \) are also \( \mathbb{R} \). However, for \( \frac{f}{g} \), we must exclude \( x \) values that make \( g(x) = 0 \). Solving \( 2x^2 - x = 0 \) gives \( x(2x - 1) = 0 \), so \( x = 0 \) or \( x = \frac{1}{2} \) must be excluded. The domain of \( \frac{f}{g} \) is \( \mathbb{R} \setminus \{0, \frac{1}{2}\} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Addition
When you add two functions, you simply add their expressions together. In this exercise, we have:
  • The function \( f(x) = x^2 + 1 \)
  • The function \( g(x) = 2x^2 - x \)
To find the sum \( (f+g)(x) \), add like terms from both functions:\[(f+g)(x) = (x^2 + 1) + (2x^2 - x) = 3x^2 - x + 1\]This expression represents the new function obtained by adding both \( f \) and \( g \).
This operation is straightforward, as you deal with each term separately to find the combined result. Adding functions is just about combining their outputs.
Function Subtraction
Subtracting one function from another involves reversing the signs of the second function's terms before combining. Given:
  • \( f(x) = x^2 + 1 \)
  • \( g(x) = 2x^2 - x \)
Find the difference \( (f-g)(x) \) by subtracting the terms:\[(f-g)(x) = (x^2 + 1) - (2x^2 - x) = x^2 + 1 - 2x^2 + x = -x^2 + x + 1\]Here, each term in \( g(x) \) is subtracted from the corresponding term in \( f(x) \).
It's important to distribute the negative sign across the \( g(x) \) terms effectively before combining similar terms.
Function Multiplication
Multiplying functions requires multiplying each term of one function by every term of the other function. Let's explore \( f(x) = x^2 + 1 \) and \( g(x) = 2x^2 - x \).
  • First, use the distributive property to expand the product:
\[(fg)(x) = (x^2 + 1)(2x^2 - x) = x^2(2x^2 - x) + 1(2x^2 - x)\]After expanding, simplify the expression:\[2x^4 - x^3 + 2x^2 - x\]This results in a new function with combined terms. Ensure all terms are accounted for and simplified.
Function multiplication often results in a higher-degree polynomial than either of the original functions.
Function Division
Function division involves creating a fraction format where the numerator is one function and the denominator is the other. Begin with:
  • \( f(x) = x^2 + 1 \) and \( g(x) = 2x^2 - x \)
The expression \( \left(\frac{f}{g}\right)(x) \) is:\[\left(\frac{f}{g}\right)(x) = \frac{x^2 + 1}{2x^2 - x}\]To perform function division, you need to be mindful of the denominator. The division is only defined when the denominator is not zero. Solve for when \( g(x) = 0 \), find that \( x = 0 \) and \( x = \frac{1}{2} \) are the points to exclude.
This is due to these values making the denominator zero, which is undefined in division.
Function Domain
Function domains specify the set of valid inputs for which the function is defined. For individual functions \( f(x) = x^2 + 1 \) and \( g(x) = 2x^2 - x \), the domain is all real numbers \( \mathbb{R} \).
  • For function addition, subtraction, and multiplication, the domain combines the domains of both functions, which is still \( \mathbb{R} \).
  • The shared domain for these operations is typically the intersection of the domains of \( f \) and \( g \).
However, for function division, the domain is adjusted to exclude any values that make the denominator zero. Solve for \( 2x^2 - x = 0 \):
  • Factor as \( x(2x - 1) = 0 \), which implies \( x = 0 \) or \( x = \frac{1}{2} \).
  • Therefore, the domain for \( \left(\frac{f}{g}\right)(x) \) is all real numbers except \{0, \frac{1}{2}\}.
Understanding domains ensures you use functions only where they make sense, avoiding undefined or non-real results.

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Most popular questions from this chapter

A few years ago a newspaper in the Midwest reported that an escape artist was planning to jump off a bridge into the Mississippi River wearing 70 lb of chains and manacles. The newspaper article stated that the height of the bridge was \(48 \mathrm{ft}\) and predicted that the escape artist's impact velocity on hitting the water would be \(85 \mathrm{mi} / \mathrm{h}\). Assuming that he simply dropped from the bridge, then his height (in feet) and velocity (in feet/second) \(t\) seconds after jumping off the bridge are given by the functions \(s(t)=\) \(-16 t^{2}+48\) and \(v(t)=-32 t,\) respectively. Determine whether the newspaper's estimate of his impact velocity was accurate.

The function \(f\) is not one-to-one on the given domain but is one-to-one on the restricted domain (the second interval). Find the inverse of the one-to- one function and give its domain. Sketch the graph of \(f\) on the restricted domain and the graph of \(f^{-1}\) on the same coordinate axes. $$ f(x)=4 x^{2}+2,(-\infty, \infty) ;[0, \infty) $$

The given function \(f\) is one-to one. Without finding \(f^{-1}\), determine the indicated function value. $$ f(x)=x^{3}+x-1 ; f^{-1}(1) $$

In Problems \(51-54\), describe in words how the graphs of the given functions differ. [Hint: Factor and cancel.] $$ f(x)=\frac{x^{2}-9}{x-3}, \quad g(x)=\left\\{\begin{array}{ll} \frac{x^{2}-9}{x-3}, & x \neq 3 \\ 4, & x=3, \end{array} \quad h(x)=\left\\{\begin{array}{ll} \frac{x^{2}-9}{x-3}, & x \neq 3 \\ 6, & x=3 \end{array}\right.\right. $$

The given function \(f\) is one-toone. The domain and range of \(f\) is given. Find \(f^{-1}\) and give its domain and range. $$ f(x)=\frac{2}{\sqrt{x}}, \quad x>0, y>0 $$
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